step one
calculate the % of oxygen
from avogadro constant
1moles = 6.02 x 10 ^23 atoms
what about 4.33 x10^22 atoms
= ( 4.33 x 10^ 22 x 1 mole ) / 6.02 10^23= 0.0719 moles
mass= 0.0719 x16= 1.1504 g
% composition is therefore= ( 1.1504/3.25) x100 = 35.40%
step two
calculate the % composition of chrorine
100- (25.42 + 35.40)=39.18%
step 3
calculate the moles of each element
that is
Na = 25.42 /23=1.1052 moles
Cl= 39.18 /35.5=1.1037moles
O= 35.40/16= 2.2125 moles
step 4
find the mole ratio by dividing each mole by 1.1037 moles
that is
Na = 1.1052/1.1037=1.001
Cl= 1.1037/1.1037= 1
0=2.2125 = 2
therefore the empirical formula= NaClO2
Answer:
B) Add appropriate quantities of weak acid and its conjugate base to water.
C) Partially neutralize a weak acid solution by addition of a strong base.
(D) Partially neutralize a weak base solution by addition of a strong acid.
Explanation:
A buffer solution is made by a weak acid and its conjugated base or a weak base and its conjugated acid.
If you add a weak acid to water, you can adjust the pH of the buffer solution by adding a strong base.
If ypu add a weak base to water, you can adjust the pH of the buffer solution by adding a strong acid.
Answer:
it's answer 2
Explanation:
magnets can only move metal
The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus 
= 0.0293 mol
Moles bromine 6.99 g bromine
=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.
